use implicit differentiation to find \\( \\frac { d y } { d x } \\) without first solving for \\( y…

Explanation:

Step1: Differentiate both sides

Differentiate (e^{5xy}) using the chain rule and product rule. The derivative of (e^{u}) with respect to (x) is (e^{u}\cdot u'), where (u = 5xy). So, ((e^{5xy})'=e^{5xy}\cdot(5y + 5x\frac{dy}{dx})).

Differentiate (y^{8}) using the chain rule. The derivative is (8y^{7}\frac{dy}{dx}).

So we have (e^{5xy}(5y + 5x\frac{dy}{dx})=8y^{7}\frac{dy}{dx}).

Step2: Expand and collect (\frac{dy}{dx}) terms

Expand the left - hand side: (5ye^{5xy}+5xe^{5xy}\frac{dy}{dx}=8y^{7}\frac{dy}{dx}).

Move all terms with (\frac{dy}{dx}) to one side: (5xe^{5xy}\frac{dy}{dx}-8y^{7}\frac{dy}{dx}=- 5ye^{5xy}).

Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(5xe^{5xy}-8y^{7})=-5ye^{5xy}).

Step3: Solve for (\frac{dy}{dx})

(\frac{dy}{dx}=\frac{-5ye^{5xy}}{5xe^{5xy}-8y^{7}}=\frac{5ye^{5xy}}{8y^{7}-5xe^{5xy}})

Answer:

(\frac{5ye^{5xy}}{8y^{7}-5xe^{5xy}})